Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373



Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373
Textbook Problem

The heat of neutralization, ΔHneut, can be defined as the amount of heat released (or absorbed), q, per mole of acid (or base) neutralized. ΔHneut for nitric acid is -52 kJ/mol HNO3. At 27.3°C, 50.00 mL of 0.743M HNO3 is neutralized by 1.00 M Sr(OH)2 in a coffee-cup calorimeter.

(a) How many mL of Sr(OH)2 were used in the neutralization?

(b) What is the final temperature of the resulting solution? (Use the assumptions in Question 11.)

Interpretation Introduction



The volume of Sr(OH)2 required to neutralize HNO3 needs to be calculated.

Concept introduction:

Heat of neutralization is the enthalpy change accompanying the neutralization of one mole of an acid by one mole of a base to form the corresponding salt and water.


Given Information:

Volume of HNO3 = 50.0 mL

Molarity of HNO3 = 0.743 M

Molarity of Sr(OH)2 = 1.00 M


The neutralization reaction between Sr (OH)2 and HNO3 can be represented as:

Sr(OH)2 + 2HNO3  Sr(NO3)2 + 2H2O

Based on the reaction stoichiometry:

1 mole of Sr(OH)2 is required to neutralize 2 moles of HNO3

Now, Molarity (M) = moles of solute (n)Volume of solution (V)Or, n = M×Number moles of Sr(OH)2 = 1/2 (number moles of HNO3)Or,M(Sr(OH)2

Interpretation Introduction



The final temperature of the solution needs to be determined.

Concept introduction:

Chemical reactions in general involve an exchange of heat between the system (reactants and products) and the surroundings. This flow of heat can be measured using a device known as a calorimeter. Heat transfer continues until an equilibrium is reached with a defined final temperature.

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